i find it difficult [Euler line of intouch triangle]

galois · Riemann Hypothesis Offline Joined: 11 Jul 2003 Posts: 397

Given a triangle $ABC$ . Let $M$ , $N$ , $P$ be the points where the incircle of triangle $ABC$ touches its sides $BC$ , $CA$ , $AB$ , respectively. Prove that the orthocenter of triangle $MNP$ (this is the point where the altitudes of triangle $MNP$ concur), the incenter of triangle $ABC$ and the circumcenter of triangle $ABC$ are collinear.

**Posted:** Tue Jun 15, 2004 8:12 pm

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vinoth_90_2004

this problem is very nice
(replace M,N,P by A',B',C' just bcoz im too lazy too remember which corresponds to which). Drop perpendiculars from H to BC, CA and AB, meeting at A'', B'' and C'' respectively. Now i think i read somewhere in the proposed problem section the following: For any point P in ABC, drop perps from P to the sides meeting at A', B', C'. define f(P)=BA'+CB'+AC'. Then for some 2 points P and Q, f(P)=f(Q) iff PQ is parellel to the line through the incentre and circumcenter. Now using this, It would suffice to prove (with this notation) f(H)=f(I)=s. After some work, this reduces to proving (with directed segments) A'A''+B'B''+C'C''=0. I cant find a nice proof of this Sad

but we can do it with some trigonometry. It reduces to proving that
sum (cyclic) of sin a * (tan b / tan c) * sin ( c - b) = 0, where a = 1/2 <CAB, b=1/2 <CBA, etc. this isnt too hard (i dont feel like posting it for some reason Mr. Green

).

**Posted:** Tue Jun 15, 2004 10:04 pm

sam-n

oh vinoth_90_2004 , you use the lemma which its proof is really harder than the orginal problem. this problem is easily solvable with homothety:

assume that the angel bisector of A and B and C intersect the circumcircle at M' and N' and P'. it is trivial that P'M'||PM and you can conclude that $H_T^k(\Delta PMN)=\Delta P'M'N'$ s.t T is a center of homothety and k is its ratio. we know that $H_T^k(H_ {\Delta PMN})=H_{\Delta P'M'N'}$ and $O_{\Delta P'M'N'}=O ,O_{\Delta PMN}=I$ so O,I,T are collinear also $H_{\Delta P'M'N'}=I,H_{\Delta PMN}=H'$ so I ,H' ,T are collinear so O,H',I are collinear.

this solves the problem. also i have seen another proof with inversion.

**Posted:** Tue Jun 15, 2004 10:47 pm

vinoth_90_2004

hmm the lemma isn't hard to prove in this case - i.e. any point K on IO satisfies f(K)=s. Obviously f(I)=f(O)=s. Since K is on O, after some ratio calculations it follows that f(K)=c1*f(I) + c2*f(O) for some constants c1,c2, c1+c2=1 and the conclusion follows. (in the general case i suppose it may be much harder tho - but we only need this case Smile

).

grobber

Here's another approach:

Fix the circumcircle and the incircle and move $A$ around on the circumcircle. The midpoint of $NP$ is the inverse of $A$ wrt the incircle, meaning that the locus of the midpoint of $NP$ is a circle (the inverse of the circumcircle wrt the incircle). This means that as we move $A$ along the circumcircle, the nine-point circle of $MNP$ is invariant, so its center is also invariant, but its center is the midpoint of the segment formed by $I$ and the orthocenter of $MNP$ , so the orthocenter of $MNP$ is also invariant.. Now take $A$ in the intersection of $IO$ and the fixed circumcircle of $ABC$ . We will have $AB=AC,\ MN=MP$ , so the orthocenter of $MNP$ will be on $OI$ , but this means that it's always there, Q.E.D.

You might have noticed that I've used a particular case of Poncelet's theorem: Given two circles $(O)$ and $(I)$ , if $(I)$ is the incircle of a triangle $ABC$ inscribed in $(O)$ , then no matter where we choose $A$ on $(O)$ , if the tangents from $A$ to $(I)$ cut $(O)$ in $B,\ C$ then $BC$ is also tangent to $(I)$ . This is a really famous result and I won't prove it here. I'm sure it can be used in contests without proof (it was even needed in one of the Mathlinks contests, but I proved it there just to make sure I get the points:)).

**Posted:** Wed Jun 16, 2004 1:09 am

darij grinberg

I remember having solved the above problem for the olympiad 2002 project (Bulgaria 35). I use slightly different notations.

Problem. Let I be the incenter of a non-equilateral triangle ABC and X, Y, Z be the tangency points of the incircle with the sides BC, CA, AB, respectively. Prove that the orthocenter of the triangle XYZ lies on the line OI, where O is the circumcenter of the triangle ABC.

Solution. Let H be the orthocenter of triangle XYZ. We want to show that H lies on OI.

Let D, E, F be the feet of the altitudes of triangle XYZ. The points E and F lie on the circle with diameter YZ; hence, < ZYF = 180° - < ZEF, or, in other words, < ZYX = < XEF. But as a secant-chord angle, < BXZ = < ZYX. Thus, < BXZ = < XEF, and EF || BC. Similarly, FD || CA and DE || AB; hence, the triangles DEF and ABC are homothetic. Let the center of homothety be S. There is a homothety with center S mapping triangle DEF to triangle ABC. This homothety must map the incenter of triangle DEF to the incenter of triangle ABC. But the incenter of triangle DEF is the orthocenter H of triangle XYZ (since triangle DEF is the orthic triangle of triangle XYZ, but the orthocenter of any acute triangle is the incenter of its orthic triangle), and the incenter of triangle ABC is I. Hence, our homothety maps H to I; therefore, S, H and I are collinear. Now, H is the orthocenter and I is the circumcenter of triangle XYZ; hence, the line HI is the Euler line of triangle XYZ. For this reason, we can say that S lies on the Euler line of triangle XYZ.

On the other hand, our homothety maps the circumcenter of triangle DEF to the circumcenter of triangle ABC. The circumcenter G of triangle DEF is the center of the nine-point circle of triangle XYZ (since triangle DEF is the orthic triangle of triangle XYZ, but the nine-point circle of any triangle is the circumcircle of its orthic triangle), and the circumcenter of triangle ABC is O. Hence, our homothety maps G to O, and consequently, S, G and O are collinear. But S lies on the Euler line of triangle XYZ (as we know), and G also lies on the Euler line of triangle XYZ (since G is the nine-point center of triangle XYZ). Hence, O must also lie on the Euler line of triangle XYZ. In other words, O lies on the line HI, and H lies on OI, qed..

This solution used homothety as the basic idea; other solutions use inversion instead.

Darij

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wellknown · P versus NP Offline Joined: 10 Jun 2004 Posts: 23

does anyone happen to know an elementary proof of Poncelet's porism for triangles ? i mean without using inversion.

**Posted:** Wed Jun 16, 2004 5:39 am

treegoner

I just want to show another idea, which is a little bit different from Darij :
Let X', Y', Z' are midpoints of YZ, ZX, XY. Thus I, X' A are collinear.
Let W is the circumcircle of X'Y'Z'.Now using the inversion $(I, r^2)$ , it will transform X' to A, Y' to B, Z' to C. Hence I, W, O are collinear.
On the other hand, IW is the Euler line of XYZ. Therefore, the orthocenter of XYZ must lie on OI.

**Posted:** Fri Jul 02, 2004 3:22 pm

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darij grinberg

Indeed. Note that in your solution, you use the

Lemma. If a circle with center O is inverted in a circle with center M, then the image is a circle whose center lies on the line MO.

I'd bet 50% of all IMO participants don't know this one ;-)

Darij

**Posted:** Sat Jul 03, 2004 12:17 am

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Sailor

I doubt. Almost everyone from Moldova solved this problem using your lemma.

**Posted:** Wed Sep 15, 2004 3:40 am

darij grinberg

No wonder that Moldova was one of the better-placed countries on the IMO 2004. But here in Germany, the lemma is pretty unknown. Most people know that an inversion maps circles to circles, and that the center usually doesn't get mapped to the corresponding center of the image circle, but they don't know about the collinearity.

Darij

**Posted:** Wed Sep 15, 2004 5:44 am

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paul_mathematics · Hodge Conjecture Offline Joined: 17 Mar 2004 Posts: 58

**Posted:** Tue Jan 25, 2005 6:40 am

darij grinberg

**Posted:** Tue Jan 25, 2005 9:12 am

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Visions of the past, dreams forsaken forming right under our eyes
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armpist · Riemann Hypothesis Offline Joined: 04 Sep 2005 Posts: 266

Make points O,I, H the intersection points of sides of an inscribed hexagon

on a Pascal line.

T.Y.

M.T.

**Posted:** Fri Feb 17, 2006 2:23 pm

pohoatza

We will use absolute barycentric coordinates. Firstly, consider $G^{\prime}$ , $O^{\prime}$ the centroid, respectively the circumcenter of the triangle $MNP$ and $O$ , $I$ the circumcenter, respectively the incenter of $ABC$ . Since

$G^{\prime} = \left(\frac {1}{3}(\frac {p - c}{b} + \frac {p - b}{c}) \ : \frac {1}{3}(\frac {p - a}{c} + \frac {p - c}{a}) \ ...$
$O^{\prime}=I= \left(\frac{a}{2p} \ : \frac{b}{2p} \ : \frac{c}{2p}\right),$
$O = \left(\frac{a^{2}(b^{2} + c^{2} - a^{2})}{2\sum{a^{2}b^{2}} - \sum{a^{4}}}} \ : \frac{b^{2}(c^{2} + a^{2} - b^{2})}{2\sum...$
the collinearity of the centroid of $MNP$ with $O$ and $I$ , is equivalent to proving that

$\left|\begin{array}{ccc}a & b & c \\ \\ a\cos A & b\cos B & c\cos C \\ \\ a[b(p - b) + c(p - c)] & b[c(...$

$\Longleftrightarrow\left|\begin{array}{ccc}1 & 1 & 1 \\ \\ \cos A & \cos B & \cos C \\ \\ b(p - b) + c(p - ...$

$\Longleftrightarrow\left|\begin{array}{ccc}1 & 1 & 1 \\ \\ a(b^{2} + c^{2} - a^{2}) & b(c^{2} + a^{2} - b^{2}) &...$

which is true. Hence, $G^{\prime}$ lies on $OI$ and since $IG^{\prime}$ is the Euler line of $MNP$ , which also contains its orthocenter, we deduce that the orthocenter lies also on the line $OI$ .

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andyciup

Consider an inversion that keeps the incircle of triangle $ABC$ as the circle of inversion.
The points $A,B,C$ become the midpoints of the triangle $MNP$ , and thus the circumcircle of triangle $ABC$ becomes the nine-point circle of triangle $MNP$ . This means that the circumcenter of triangle $ABC$ becomes the nine-point-center for triangle $MNP$ . But because our inversion is of center $I$ , this means that the circumcenter of ABC, the circumcenter of ABC after the inversion, and the incircle of triangle $ABC$ are coliniar, which is equivalent to saying that the circumcenter of $ABC$ belongs to the line conecting the circumcenter of $MNP$ with the nine-point center of $MNP$ , that is, the Euler line of MNP.
QED.

**Posted:** Sun Apr 22, 2007 10:50 am

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edriv

**Posted:** Sun Apr 22, 2007 11:54 am

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**Posted:** Tue Apr 24, 2007 3:47 pm

gemath · Yang-Mills Theory Offline Joined: 17 Mar 2006 Posts: 979

I think this problem was well know. A proof using vectors:
The orthocenter $H^{\prime}$ of triangle $MNP$ lies on Euler line of triangle $MNP$ , but the Euler line of $MNP$ is line $OI$ . Indeed let intersection of lines $IA,IB,IC$ with $(O)$ be $A',B',C'$ easily seen $I$ is orthocenter of $A'B'C'\Rightarrow \vec{OI}=\vec{OA'}+\vec{OB'}+\vec{OC'}$ but easily seen $\vec{OA'}=\frac{R}{r}\vec{IM}\Rightarrow \vec{OI}=\frac{R}{r}(\vec{IM}+\vec{IN}+\vec{IP})=\frac{R}{r}\vec{IH^{\prime}}$
Thus we not only show $H^{\prime}\in OI$ but also we show $IH^{\prime}=\frac{r}{R}OI$ nice result Mr. Green

**Posted:** Wed Apr 25, 2007 2:34 am

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Virgil Nicula

Lemma. Let $ABC$ be a triangle with the orthocenter $H$ and the circumcircle $C(O)$ . Then there is the Sylvester's identity
$\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC}=2\cdot\overrightarrow{HO}$ which is equivalently with $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OH}$ .

Remark. If we"ll choose $O$ as the origin of the vectorial system, i.e. $X\equiv \overrightarrow{OX}$ and $\overrightarrow{XY}=Y-X$ , then Sylvester's identity becomes $\boxed{A+B+C=H}$ .

Proof III (Gemath). Denote the second intersections $A',B',C'$ of the lines $IA,IB,IC$ with the circumcircle of $\triangle ABC$ . Prove easily that the incenter $I$ of $\triangle ABC$ is the orthocenter of $\triangle A'B'C'$ , i.e. $A'+B'+C'=I$ . From the evident relations $\frac{OA'}{ID}=\frac{R}{r}$ and $OA'\parallel ID$ obtain $r\cdot\overrightarrow{OA'}=R\cdot\overrightarrow{ID}$ , i.e. $r\cdot A'=R\cdot (D-I)$ a.s.o. Therefore, $r\cdot I=r\cdot (A'+B'+C')=R\cdot (D+E+F-3\cdot I)=3R\cdot (G'-I)$ , i.e. $r\cdot \overrightarrow{OI}=3R\cdot \overrightarrow{IG'}$ ., what means $G'\in OI$ . If $H'$ is the orthocenter of $\triangle DEF$ , then $3\cdot \overrightarrow{IG'}=\overrightarrow{IH'}$ (the line $\overline{H'G'I}$ is the Euler's line for $\triangle DEF$ ) and the previous relation becomes $\boxed{\ r\cdot \overrightarrow{OI}=R\cdot \overrightarrow{IH'}\ }$ .

**Posted:** Wed Apr 25, 2007 6:07 pm